Evaluate
`int(1+2x^(2))/(x^(2)(1+x^(2)))dx`

Answer

To evaluate the integral

\[ \int \frac{1 + 2x^2}{x^2(1 + x^2)} \, dx, \]

we can follow these steps:

### Step 1: Simplify the integrand
We start with the expression

\[ \frac{1 + 2x^2}{x^2(1 + x^2)}. \]

We can rewrite \(1 + 2x^2\) as \(1 + x^2 + x^2\). This allows us to break it down:

\[ 1 + 2x^2 = (1 + x^2) + x^2. \]

Now, we can express the integrand as:

\[ \frac{(1 + x^2) + x^2}{x^2(1 + x^2)} = \frac{1 + x^2}{x^2(1 + x^2)} + \frac{x^2}{x^2(1 + x^2)}. \]

### Step 2: Separate the fractions
This can be separated into two integrals:

\[ \int \left( \frac{1 + x^2}{x^2(1 + x^2)} + \frac{x^2}{x^2(1 + x^2)} \right) \, dx = \int \frac{1}{x^2} \, dx + \int \frac{1}{1 + x^2} \, dx. \]

### Step 3: Evaluate the integrals
Now we evaluate each integral separately.

1. For the first integral:

\[ \int \frac{1}{x^2} \, dx = \int x^{-2} \, dx = -\frac{1}{x} + C_1. \]

2. For the second integral:

\[ \int \frac{1}{1 + x^2} \, dx = \tan^{-1}(x) + C_2. \]

### Step 4: Combine the results
Combining the results from both integrals, we have:

\[ -\frac{1}{x} + \tan^{-1}(x) + C, \]

where \(C = C_1 + C_2\) is the constant of integration.

### Final Answer
Thus, the final answer is:

\[ \int \frac{1 + 2x^2}{x^2(1 + x^2)} \, dx = -\frac{1}{x} + \tan^{-1}(x) + C. \]